1. Chapter 6
Section 4

2. A General Approach to Factoring
6.4
A General Approach to Factoring
Factor out any common factor.
Factor binomials.
Factor trinomials.
Factor polynomials of more than three terms.

3. Factoring a Polynomial
A General Approach to Factoring
Factoring a Polynomial
Step 1 Factor out any common factor.
Step 2 If the polynomial is a binomial, check to see if it is the difference of squares, a difference of cubes, or a sum of cubes.
If the polynomial is a trinomial, check to see if it is a perfect square trinomial. If it is not, factor as in Section 6.2.
If the polynomial has more than three terms, try to factor by grouping.
Step 3 Check the factored form by multiplying.
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4. Factor out any common factor.
Objective 1
Factor out any common factor.
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5. Factoring Out a Common Factor
CLASSROOM EXAMPLE 1
Factoring Out a Common Factor
Factor each polynomial.
2x3 + 10x2 – 4x
The GCF is 2x.
= 2x(x2 + 5x – 2)
12m(p – q) – 7n(p – q)
The GCF is (p – q)
= (p – q)(12m – 7n)
Solution:
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6. Objective 2
Factor binomials.
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7. Factor binomials. Factoring a Binomial
For a binomial (two terms), check for the following:
Difference of Squares:
x2 – y2 = (x – y)(x + y)
Difference of Cubes:
x3 – y3 = (x – y)(x2 + xy + y2)
Sum of Cubes:
x3 + y3 = (x + y)(x2 – xy + y2)
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8. Factor each binomial, if possible. 36x2 – y2 Difference of squares
CLASSROOM EXAMPLE 2
Factoring Binomials
Factor each binomial, if possible.
36x2 – y2
Difference of squares
= (6x)2 – (y)2
= (6x – y)(6x + y)
4t2 + 1
The binomial is prime. It is the sum of squares.
125x3 – 27y3 =
= (5x – 3y)[(5x)2 + (5x)(3y) + (3y)2]
= (5x – 3y)(25x2 + 15xy + 9y2)
Solution:
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9. Objective 3
Factor trinomials.
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10. Factor trinomials. Factoring a Trinomial
For a trinomial (three terms), decide whether it is a perfect square trinomial of either of these forms
x2 + 2xy + y2 = (x + y)2 or
x2 – 2xy + y2 = (x – y)2
If not, use the methods of Section 6.2.
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11. Perfect square trinomial
CLASSROOM EXAMPLE 3
Factoring Trinomials
Factor each trinomial.
16m2 + 56m + 49
= (4m + 7)2
8t2 – 13t + 5
Two integer factors whose sum is 8(5) = 40 and whose sum is –13 are –5 and –8.
= 8t2 – 5t – 8t + 5
= t(8t – 5) – 1(8t – 5)
= (8t – 5)(t – 1)
Solution:
Perfect square trinomial
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12. Factoring Trinomials (cont’d)
CLASSROOM EXAMPLE 3
Factoring Trinomials (cont’d)
Factor the trinomial.
6x2 – 3x – 63
Factor out the GCF of 3.
= 3(2x2 – x – 21)
Two factors whose product is 2(–21) = – 42 and whose sum is –1 are –7 and 6.
= 3[2x2 – 7x + 6x – 21]
= 3[x(2x – 7) + 3(2x – 7)]
= 3(2x – 7)(x + 3)
Solution:
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13. Factor polynomials of more than three terms.
Objective 3
Factor polynomials of more than three terms.
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14. Factoring Polynomials with More than Three Terms
CLASSROOM EXAMPLE 4
Factoring Polynomials with More than Three Terms
Factor each polynomial.
p3 – 2pq2 + p2q – 2q3
= (p3 – 2pq2) +(p2q – 2q3)
= p(p2 – 2q2) + q(p2 – 2q2)
= (p2 – 2q2)(p + q)
9x2 + 24x + 16 – y2
= (9x2 + 24x + 16) – y2
= (3x + 4)2 – y2
= [(3x + 4) + y)][(3x + 4) – y)]
= (3x y)(3x + 4 – y)
Solution:
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15. = {(4a + b)[(4a)2 – 4ab + b2]} + [(4a + b)(4a – b)]
CLASSROOM EXAMPLE 4
Factoring Polynomials with More than Three Terms (cont’d)
Factor the polynomial.
64a3 + 16a2 + b3 – b2
= (64a3 + b3) + (16a2 – b2)
= [(4a)3 + b3] + [(4a)2 – b2]
= {(4a + b)[(4a)2 – 4ab + b2]} + [(4a + b)(4a – b)]
= [(4a + b)(16a2 – 4ab + b2)] + [(4a + b)(4a – b)]
= (4a + b)(16a2 – 4ab + b2 + 4a – b)
Solution:
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